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HLIBMathematics

Mathematics: Analysis and Approaches

The Higher Level pathway extends the AA SL foundations through advanced algebra, functions, geometry, probability and calculus.

Course units

5

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Not specified

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Course Overview

IB Mathematics: Analysis and Approaches HL is organized into 5 units and 83 focused topics with 1311 Key Concepts sections, 1048 questions, and 1660 flashcards.

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Begin with Topic 1.1: Scientific notation and operations with numbers.

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Your learning path

Course units

Each unit groups focused topics. Every topic follows the same clear rhythm: review Key Concepts, use the Question Bank, then strengthen recall.

  1. Unit 1
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    Number and Algebra

    AA HL includes the common SL foundations in number and algebra, then extends them through counting, generalized binomial expansions, partial fractions, complex numbers, advanced proof and linear systems.

    16 topics

    Unit 1 progress0%

    Topics in this unit

    Topic 1.1Not started

    Scientific notation and operations with numbers

    Scientific notation is a compact, reliable way to represent, compare and calculate with extremely large or small numbers. This guide builds from place value and conversions to arithmetic, significant figures, calculator use and IB-style multi-step problems.

    105 min estimated

    Topic 1.1 progress0%
    Learning objectives
    • Write every non-zero number in normalized scientific notation a × 10k10^{k}, where 1 ≤ |a| < 10 and k is an integer.
    • Convert accurately between ordinary decimal notation, scientific notation and calculator E-notation.
    • Multiply, divide and raise scientific-notation values to integer powers using exponent laws.
    • Add and subtract scientific-notation values by first expressing them with a common power of ten.
    • Round values to a stated number of significant figures and communicate trailing significant zeros clearly.
    • Use a scientific calculator efficiently while retaining guard digits until the final answer.
    • Estimate orders of magnitude and use them to check whether a calculated result is reasonable.
    • Present IB exam solutions with correct working, notation, accuracy and units.
    Key formulas

    Normalized scientific notation

    N=a×10k,1≤∣a∣<10,k∈ZN=a\times10^k,\quad 1\le |a|<10,\quad k\in\mathbb Z

    Multiplication rule

    (a×10m)(b×10n)=(ab)×10m+n(a\times10^m)(b\times10^n)=(ab)\times10^{m+n}

    Division rule

    a×10mb×10n=ab×10m−n,b≠0\frac{a\times10^m}{b\times10^n}=\frac{a}{b}\times10^{m-n},\quad b\ne0

    Power rule

    (a×10m)p=ap×10mp(a\times10^m)^p=a^p\times10^{mp}

    Addition with a common exponent

    a×10k+b×10k=(a+b)×10ka\times10^k+b\times10^k=(a+b)\times10^k

    Subtraction with a common exponent

    a×10k−b×10k=(a−b)×10ka\times10^k-b\times10^k=(a-b)\times10^k

    Calculator E-notation

    aEk=a×10ka\text{E}k=a\times10^k
    Topic 1.2Not started

    Arithmetic sequences and series

    Arithmetic sequences model quantities that change by the same additive amount each step. This guide develops common differences, nth terms, finite sums, sigma notation and IB-style applications from first principles.

    110 min estimated

    Topic 1.2 progress0%
    Learning objectives
    • Recognize an arithmetic sequence and calculate its common difference from consecutive or non-consecutive terms.
    • Use recurrence notation and the nth-term formula to generate and find terms.
    • Determine an unknown first term, common difference, term number or number of terms from given information.
    • Distinguish an arithmetic sequence from an arithmetic series and calculate the sum of a finite number of terms.
    • Represent and evaluate finite arithmetic sums using sigma notation.
    • Build and interpret arithmetic models for real-world situations with constant additive change.
    • Use calculator tables, summation tools and equation solvers efficiently while checking integer constraints.
    • Present IB exam solutions with clear formulas, substitutions, units and reasonableness checks.
    Key formulas

    Common difference

    d=un+1−und=u_{n+1}-u_n

    Recursive definition

    u1=a,un+1=un+du_1=a,\qquad u_{n+1}=u_n+d

    Arithmetic nth term

    un=u1+(n−1)du_n=u_1+(n-1)d

    Common difference from two terms

    d=uq−upq−p,p≠qd=\frac{u_q-u_p}{q-p},\qquad p\ne q

    Term number

    n=1+un−u1d,d≠0n=1+\frac{u_n-u_1}{d},\qquad d\ne0

    Arithmetic series using first and last terms

    Sn=n2(u1+un)S_n=\frac n2(u_1+u_n)

    Arithmetic series using the common difference

    Sn=n2[2u1+(n−1)d]S_n=\frac n2\left[2u_1+(n-1)d\right]

    Arithmetic series in sigma notation

    Sn=∑r=1n[u1+(r−1)d]S_n=\sum\limits_{r=1}^{n}\left[u_1+(r-1)d\right]

    Equivalent zero-based indexing

    Sn=∑r=0n−1(u1+rd)S_n=\sum\limits_{r=0}^{n-1}(u_1+rd)
    Topic 1.3Not started

    Geometric sequences and series

    Geometric sequences change by a constant ratio. AA HL questions may connect geometric growth or decay to repeated percentage change, modelling and series notation.

    120 min estimated

    Topic 1.3 progress0%
    Learning objectives
    • Identify the common ratio and construct terms.
    • Use the nth-term formula and finite geometric sum.
    • Recognize when a geometric model is appropriate in context.
    Key formulas

    Common ratio

    r=un+1un,un≠0r=\frac{u_{n+1}}{u_n},\qquad u_n\ne0

    Recursive definition

    u1=a,un+1=runu_1=a,\qquad u_{n+1}=ru_n

    Geometric nth term

    un=u1rn−1u_n=u_1r^{n-1}

    Ratio from two terms

    rq−p=uqupr^{q-p}=\frac{u_q}{u_p}

    Term number for a positive ratio

    n=1+ln⁡(unu1)ln⁡r,r>0, r≠1,u1≠0,unu1>0n=1+\frac{\ln(\frac{u_n}{u_1})}{\ln r},\qquad r>0,\ r\ne1,\quad u_1\ne0,\quad \frac{u_n}{u_1}>0

    Finite geometric sum

    Sn=u1(1−rn)1−r,r≠1S_n=\frac{u_1(1-r^n)}{1-r},\qquad r\ne1

    Equivalent finite-sum form

    Sn=u1(rn−1)r−1,r≠1S_n=\frac{u_1(r^n-1)}{r-1},\qquad r\ne1

    Special case r = 1

    Sn=nu1S_n=nu_1

    Infinite geometric sum

    S∞=u11−r,∣r∣<1S_\infty=\frac{u_1}{1-r},\qquad |r|<1

    Remainder after n terms

    Rn=S∞−Sn=u1rn1−r,∣r∣<1R_n=S_\infty-S_n=\frac{u_1r^n}{1-r},\qquad |r|<1

    Finite geometric series in sigma notation

    Sn=∑k=0n−1u1rkS_n=\sum\limits_{k=0}^{n-1}u_1r^k

    Infinite geometric series in sigma notation

    ∑k=0∞u1rk=u11−r,∣r∣<1\sum\limits_{k=0}^{\infty}u_1r^k=\frac{u_1}{1-r},\qquad |r|<1

    Repeated growth or decay

    Qt=Q0rtQ_t=Q_0r^t
    Topic 1.4Not started

    Financial applications (current syllabus)

    Financial mathematics applies exponential growth and decay to money over repeated periods. Students should understand how percentage rate, number of compounding periods and time units interact.

    140 min estimated

    Topic 1.4 progress0%
    Learning objectives
    • Model compound interest and repeated percentage growth.
    • Model depreciation and interpret the value after a given number of periods.
    • Translate annual, monthly or other periodic information consistently before substituting.
    Key formulas

    Simple interest earned

    I=PrtI=Prt

    Simple-interest balance

    A=P(1+rt)A=P(1+rt)

    Percentage multiplier

    growth: 1+p100,decay: 1−p100\text{growth: }1+\frac{p}{100},\qquad \text{decay: }1-\frac{p}{100}

    Compound future value

    FV=PV(1+i)nFV=PV(1+i)^n

    Periodic rate from two values

    i=(FVPV)1n−1,n>0, PV≠0, FVPV>0i=\left(\frac{FV}{PV}\right)^{\frac{1}{n}}-1,\quad n>0,\ PV\ne0,\frac{\ FV}{PV}>0

    Number of periods

    n=ln⁡(FV/PV)ln⁡(1+i)PV≠0,FV/PV>0,i>−1, i≠0\begin{gathered}n=\frac{\ln(FV/PV)}{\ln(1+i)}\\ PV\ne0,\quad FV/PV>0,\quad i>-1,\ i\ne0\end{gathered}

    Nominal-rate compound interest

    FV=PV(1+rm)mtFV=PV\left(1+\frac{r}{m}\right)^{mt}

    Periodic rate and period count

    i=rm,n=mti=\frac{r}{m},\qquad n=mt

    Effective annual rate

    EAR=(1+rm)m−1EAR=\left(1+\frac{r}{m}\right)^m-1

    Present value of one future amount

    PV=FV(1+i)n=FV(1+i)−nPV=\frac{FV}{(1+i)^n}=FV(1+i)^{-n}

    Appreciation

    Vn=V0(1+a)nV_n=V_0(1+a)^n

    Reducing-balance depreciation

    Vn=V0(1−d)nV_n=V_0(1-d)^n

    Future value of an ordinary annuity

    FV={R(1+i)n−1i,i≠0nR,i=0FV=\begin{cases}R\frac{(1+i)^n-1}{i},&i\ne0\\ nR,&i=0\end{cases}

    Future value of an annuity due

    FVdue={R(1+i)n−1i(1+i),i≠0nR,i=0FV_{\mathrm{due}}=\begin{cases}R\frac{(1+i)^n-1}{i}(1+i),&i\ne0\\ nR,&i=0\end{cases}

    Deposit required for a target

    R={FVi(1+i)n−1,i≠0FV/n,i=0R=\begin{cases}FV\frac{i}{(1+i)^n-1},&i\ne0\\ FV/n,&i=0\end{cases}

    Present value of loan repayments

    PV={R1−(1+i)−ni,i≠0nR,i=0PV=\begin{cases}R\frac{1-(1+i)^{-n}}{i},&i\ne0\\ nR,&i=0\end{cases}

    Periodic loan repayment

    R={PVi1−(1+i)−n,i≠0PV/n,i=0R=\begin{cases}PV\frac{i}{1-(1+i)^{-n}},&i\ne0\\ PV/n,&i=0\end{cases}

    Outstanding balance after k payments

    Bk={PV(1+i)k−R(1+i)k−1i,i≠0PV−kR,i=0B_k=\begin{cases}PV(1+i)^k-R\frac{(1+i)^k-1}{i},&i\ne0\\ PV-kR,&i=0\end{cases}

    Total interest on equal repayments

    Itotal=Rn−PVI_{\mathrm{total}}=Rn-PV
    Topic 1.5Not started

    Integer exponents and introduction to logarithms

    This section links exponent rules to logarithms and introduces logarithms as inverse operations. Technology can be used for numerical evaluation, but students should understand what a logarithm represents.

    125 min estimated

    Topic 1.5 progress0%
    Learning objectives
    • Apply the laws of exponents for integer powers.
    • Evaluate common logarithms and natural logarithms numerically.
    • Connect exponential equations with logarithmic notation.
    Key formulas

    Positive integer exponent

    an=a⋅a⋯a⏟n factors,n∈Na^n=\underbrace{a\cdot a\cdots a}_{n\text{ factors}},\quad n\in\mathbb N

    Zero exponent

    a0=1,a≠0a^0=1,\quad a\ne0

    Negative integer exponent

    a−n=1an,a≠0a^{-n}=\frac{1}{a^n},\quad a\ne0

    Product law

    aman=am+na^m a^n=a^{m+n}

    Quotient law

    aman=am−n,a≠0\frac{a^m}{a^n}=a^{m-n},\quad a\ne0

    Power of a power

    (am)n=amn(a^m)^n=a^{mn}

    Power of a product

    (ab)n=anbn(ab)^n=a^n b^n

    Power of a quotient

    (ab)n=anbn,b≠0\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n},\quad b\ne0

    Moving a negative power across a fraction bar

    a−mb−n=bnam,a,b≠0\frac{a^{-m}}{b^{-n}}=\frac{b^n}{a^m},\quad a,b\ne0

    Exponential-logarithmic equivalence

    ac=b⟺log⁡ab=ca^c=b\quad\Longleftrightarrow\quad\log_a b=c

    Common and natural logarithms

    log⁡x=log⁡10x,ln⁡x=log⁡ex\log x=\log_{10}x,\qquad\ln x=\log_e x

    Inverse identities

    log⁡a(ax)=x,alog⁡ax=x(x>0)\log_a(a^x)=x,\qquad a^{\log_a x}=x\quad(x>0)

    Product property

    log⁡a(xy)=log⁡ax+log⁡ay\log_a(xy)=\log_a x+\log_a y

    Quotient property

    log⁡a(xy)=log⁡ax−log⁡ay\log_a\left(\frac{x}{y}\right)=\log_a x-\log_a y

    Power property

    log⁡a(xk)=klog⁡ax\log_a(x^k)=k\log_a x

    Useful exact values

    log⁡a1=0,log⁡aa=1\log_a1=0,\qquad\log_a a=1

    Common-base principle

    af(x)=ag(x)⟹f(x)=g(x),a>0, a≠1a^{f(x)}=a^{g(x)}\Longrightarrow f(x)=g(x),\quad a>0,\ a\ne1

    Solving a basic exponential equation

    ax=b⟹x=ln⁡bln⁡a,a>0, a≠1, b>0a^x=b\Longrightarrow x=\frac{\ln b}{\ln a},\quad a>0,\ a\ne1,\ b>0

    Single logarithm equation

    log⁡af(x)=c⟺f(x)=ac,f(x)>0\log_a f(x)=c\Longleftrightarrow f(x)=a^c,\quad f(x)>0

    Equality of logarithms

    log⁡af(x)=log⁡ag(x)⟹f(x)=g(x),f(x),g(x)>0\log_a f(x)=\log_a g(x)\Longrightarrow f(x)=g(x),\quad f(x),g(x)>0

    Change-of-base formula

    log⁡ab=log⁡blog⁡a=ln⁡bln⁡a\log_a b=\frac{\log b}{\log a}=\frac{\ln b}{\ln a}
    Topic 1.6Not started

    Simple deductive proof

    AA HL expects students to justify mathematical statements rather than rely only on numerical evidence. Proof questions are usually short, but notation and logical flow matter.

    115 min estimated

    Topic 1.6 progress0%
    Learning objectives
    • Build a numerical or algebraic argument from accepted facts.
    • Show a clear chain of equivalences or implications.
    • Distinguish an equation that is true for particular values from an identity that is true throughout its domain.
    Key formulas

    Implication structure

    P⇒QP\Rightarrow Q

    Universal statement

    ∀n∈S,P(n)\forall n\in S,\quad P(n)

    Even integer

    n=2k,k∈Zn=2k,\quad k\in\mathbb Z

    Odd integer

    n=2k+1,k∈Zn=2k+1,\quad k\in\mathbb Z

    Divisibility

    d∣n⟺n=dk for some k∈Zd\mid n\quad\Longleftrightarrow\quad n=dk\text{ for some }k\in\mathbb Z

    Consecutive integers

    n, n+1, n+2,…(n∈Z)n,\ n+1,\ n+2,\ldots\quad(n\in\mathbb Z)

    Direct-proof chain

    P ⟹ R1 ⟹ R2 ⟹⋯⟹ QP\ \Longrightarrow\ R_1\ \Longrightarrow\ R_2\ \Longrightarrow\cdots\Longrightarrow\ Q

    Identity notation

    f(x)≡g(x)f(x)\equiv g(x)

    Difference of squares

    A2−B2=(A−B)(A+B)A^2-B^2=(A-B)(A+B)

    Square is non-negative

    u2≥0for every u∈Ru^2\ge0\quad\text{for every }u\in\mathbb R

    Even target form

    E=2K,K∈ZE=2K,\quad K\in\mathbb Z

    Odd target form

    O=2K+1,K∈ZO=2K+1,\quad K\in\mathbb Z

    Odd-square result

    n odd⟹n2≡1(mod8)n\text{ odd}\quad\Longrightarrow\quad n^2\equiv1\pmod8

    Divisibility target

    d∣N⟺N=dK for some K∈Zd\mid N\quad\Longleftrightarrow\quad N=dK\text{ for some }K\in\mathbb Z

    Linear-combination property

    d∣a ∧ d∣b⟹d∣(ra+sb),r,s∈Zd\mid a\ \land\ d\mid b\quad\Longrightarrow\quad d\mid(ra+sb),\quad r,s\in\mathbb Z

    Consecutive integers

    n, n+1, n+2n,\ n+1,\ n+2

    Consecutive even integers

    2k, 2k+2, 2k+42k,\ 2k+2,\ 2k+4

    Consecutive odd integers

    2k+1, 2k+3, 2k+52k+1,\ 2k+3,\ 2k+5

    Counterexample test

    P(c) true and Q(c) false⟹∀x [P(x)⇒Q(x)] is falseP(c)\text{ true and }Q(c)\text{ false}\quad\Longrightarrow\quad \forall x\,[P(x)\Rightarrow Q(x)]\text{ is false}
    Topic 1.7Not started

    Rational exponents and logarithm laws

    Students extend exponent rules to rational powers and use logarithm laws to simplify expressions and solve exponential equations. This topic appears frequently inside function and calculus questions as well.

    125 min estimated

    Topic 1.7 progress0%
    Learning objectives
    • Apply product, quotient and power laws for logarithms.
    • Change the base of a logarithm when necessary.
    • Solve exponential equations by taking logarithms and check domain restrictions.
    Key formulas

    Rational exponent definition

    amn=amn=(an)ma^{\frac{m}{n}}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m

    Negative rational exponent

    a−mn=1amna^{\frac{-m}{n}}=\frac{1}{a^{\frac{m}{n}}}

    Special root form

    a1n=ana^{\frac{1}{n}}=\sqrt[n]{a}

    Radical–exponent conversion

    xmn=xmn,x>0, m∈Z, n∈Z>0\sqrt[n]{x^m}=x^{\frac{m}{n}},\qquad x>0,\ m\in\mathbb Z,\ n\in\mathbb Z_{>0}

    Principal square root

    x2=∣x∣\sqrt{x^2}=|x|

    Odd root of a power

    x2k+12k+1=x\sqrt[2k+1]{x^{2k+1}}=x

    Product law

    aras=ar+sa^r a^s=a^{r+s}

    Quotient law

    aras=ar−s\frac{a^r}{a^s}=a^{r-s}

    Power of a power

    (ar)s=ars(a^r)^s=a^{rs}

    Power of a product

    (ab)r=arbr(ab)^r=a^r b^r

    Power of a quotient

    (ab)r=arbr\left(\frac ab\right)^r=\frac{a^r}{b^r}

    Negative exponent conversion

    x−r=1xrx^{-r}=\frac{1}{x^r}

    Common-denominator strategy

    xpqxrs=x(ps+rq)(qs)x^{\frac{p}{q}}x^{\frac{r}{s}}=x^{\frac{(ps+rq)}{(qs)}}

    Exact before decimal

    amn=amn\sqrt[n]{a^m}=a^{\frac{m}{n}}

    Logarithm definition

    log⁡ax=y⟺ay=x\log_a x=y\quad\Longleftrightarrow\quad a^y=x

    Logarithm of one

    log⁡a1=0\log_a1=0

    Logarithm of its base

    log⁡aa=1\log_a a=1

    Inverse relationships

    alog⁡ax=x,log⁡a(ax)=xa^{\log_a x}=x,\qquad \log_a(a^x)=x

    Product law

    log⁡a(xy)=log⁡ax+log⁡ay\log_a(xy)=\log_a x+\log_a y

    Quotient law

    log⁡a(xy)=log⁡ax−log⁡ay\log_a\left(\frac{x}{y}\right)=\log_a x-\log_a y

    Power law

    log⁡a(xr)=rlog⁡ax\log_a(x^r)=r\log_a x

    Reciprocal result

    log⁡a(1x)=−log⁡ax\log_a\left(\frac1x\right)=-\log_a x

    Change of base

    log⁡ax=log⁡bxlog⁡ba\log_a x=\frac{\log_b x}{\log_b a}

    Calculator forms

    log⁡ax=ln⁡xln⁡a=log⁡xlog⁡a\log_a x=\frac{\ln x}{\ln a}=\frac{\log x}{\log a}

    Condensing pattern

    rlog⁡ax−slog⁡ay=log⁡a(xrys)r\log_a x-s\log_a y=\log_a\left(\frac{x^r}{y^s}\right)

    Same-base principle

    au=av⟹u=va^u=a^v\quad\Longrightarrow\quad u=v

    Logarithmic solution step

    af(x)=k⟹f(x)=ln⁡kln⁡aa^{f(x)}=k\quad\Longrightarrow\quad f(x)=\frac{\ln k}{\ln a}

    Log equality principle

    log⁡au=log⁡av⟹u=v\log_a u=\log_a v\quad\Longrightarrow\quad u=v

    Fractional exponent

    amn=amna^{\frac{m}{n}}=\sqrt[n]{a^m}

    Three log laws

    log⁡a(xy)=log⁡ax+log⁡ay,log⁡a(xy)=log⁡ax−log⁡ay,log⁡a(xr)=rlog⁡ax\log_a(xy)=\log_ax+\log_ay,\quad \log_a(\frac{x}{y})=\log_ax-\log_ay,\quad \log_a(x^r)=r\log_ax

    Change of base

    log⁡ax=ln⁡xln⁡a\log_a x=\frac{\ln x}{\ln a}
    Topic 1.8Not started

    Infinite geometric series

    An infinite geometric series has a finite sum only when its terms shrink fast enough. Students must identify convergence before applying the sum formula.

    110 min estimated

    Topic 1.8 progress0%
    Learning objectives
    • Determine whether |r| < 1.
    • Find the limiting sum of a convergent geometric series.
    • Use an infinite series model in an applied or algebraic context.
    Key formulas

    Infinite geometric series

    a+ar+ar2+ar3+⋯a+ar+ar^2+ar^3+\cdots

    Partial sum notation

    Sn=∑k=0n−1arkS_n=\sum\limits_{k=0}^{n-1}ar^k

    Common ratio

    r=un+1unr=\frac{u_{n+1}}{u_n}

    Geometric term

    un=arn−1u_n=ar^{n-1}

    Geometric convergence condition

    ∣r∣<1⟺−1<r<1|r|<1\quad\Longleftrightarrow\quad -1<r<1

    Vanishing powers

    ∣r∣<1⟹rn→0|r|<1\quad\Longrightarrow\quad r^n\to0

    Finite geometric sum

    Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r}

    Sum to infinity

    S∞=a1−r,∣r∣<1S_\infty=\frac{a}{1-r},\qquad |r|<1

    First term from the sum

    a=S∞(1−r)a=S_\infty(1-r)

    Ratio from first term and sum

    r=1−aS∞,S∞≠0r=1-\frac{a}{S_\infty},\qquad S_\infty\ne0

    Term condition

    un=arn−1u_n=ar^{n-1}

    Remainder after n terms

    Rn=S∞−Sn=arn1−rR_n=S_\infty-S_n=\frac{ar^n}{1-r}

    Approximation relation

    S∞=Sn+RnS_\infty=S_n+R_n

    Bouncing-ball distance

    D=h+2hq1−q=h(1+q1−q)D=h+\frac{2hq}{1-q}=h\left(\frac{1+q}{1-q}\right)

    Repeated fixed contribution

    A∞=d1−r,∣r∣<1A_\infty=\frac{d}{1-r},\qquad |r|<1

    Two-digit recurring decimal

    0.ab‾=10a+b99,a,b∈{0,1,…,9}0.\overline{ab}=\frac{10a+b}{99},\qquad a,b\in\{0,1,\ldots,9\}

    Partial-sum calculator entry

    Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r}

    Solver interval

    −1<r<1-1<r<1

    Convergence and sum

    ∣r∣<1⟹S∞=a1−r|r|<1\quad\Longrightarrow\quad S_\infty=\frac{a}{1-r}

    Term and remainder

    un=arn−1,Rn=arn1−ru_n=ar^{n-1},\qquad R_n=\frac{ar^n}{1-r}
    Topic 1.9Not started

    Binomial theorem

    The binomial theorem provides a systematic expansion of powers such as (a+b)n.\left(a + b\right)^{n.} Students should be able to identify individual terms without expanding the entire expression.

    115 min estimated

    Topic 1.9 progress0%
    Learning objectives
    • Use binomial coefficients and Pascal’s triangle.
    • Find a specified term or coefficient in an expansion.
    • Connect the general term with powers of the two binomial components.
    Key formulas

    Binomial theorem

    (a+b)n=∑r=0n(nr)an−rbr(a+b)^n=\sum\limits_{r=0}^{n}\binom nr a^{n-r}b^r

    Number of formal binomial summands

    formal summands before collection=n+1\text{formal summands before collection}=n+1

    Pascal recurrence

    (nr)=(n−1r−1)+(n−1r)\binom nr=\binom{n-1}{r-1}+\binom{n-1}{r}

    Coefficient symmetry

    (nr)=(nn−r)\binom nr=\binom n{n-r}

    Row sum

    ∑r=0n(nr)=2n\sum\limits_{r=0}^{n}\binom nr=2^n

    Combination formula

    (nr)=n!r!(n−r)!\binom nr=\frac{n!}{r!(n-r)!}

    Factorial

    n!=n(n−1)(n−2)⋯2⋅1,0!=1n!=n(n-1)(n-2)\cdots2\cdot1,\qquad 0!=1

    Boundary values

    (n0)=(nn)=1\binom n0=\binom nn=1

    Expanded form

    (a+b)n=(n0)an+(n1)an−1b+⋯+(nn)bn(a+b)^n=\binom n0a^n+\binom n1a^{n-1}b+\cdots+\binom nnb^n

    Exponent check

    (n−r)+r=n(n-r)+r=n

    General term

    Tr+1=(nr)an−rbrT_{r+1}=\binom nr a^{n-r}b^r

    pth-term index

    r=p−1r=p-1

    Coefficient method

    power of x=k⟹solve for r\text{power of }x=k\quad\Longrightarrow\quad \text{solve for }r

    Quadratic coefficient

    (n2)=n(n−1)2\binom n2=\frac{n(n-1)}2

    Constant-term condition

    net exponent of x=0\text{net exponent of }x=0

    Admissible index

    r∈{0,1,2,…,n}r\in\{0,1,2,\ldots,n\}

    Sum of coefficients

    ∑k=0nck=P(1)\sum\limits_{k=0}^{n}c_k=P(1)

    Alternating coefficient sum

    ∑k=0n(−1)kck=P(−1)\sum\limits_{k=0}^{n}(-1)^kc_k=P(-1)

    GDC term calculation

    (nr)an−rbr\binom nr a^{n-r}b^r

    Core theorem

    (a+b)n=∑r=0n(nr)an−rbr(a+b)^n=\sum\limits_{r=0}^{n}\binom nr a^{n-r}b^r

    Target term

    Tr+1=(nr)an−rbrT_{r+1}=\binom nr a^{n-r}b^r

    Constant condition

    net power of x=0\text{net power of }x=0
    Topic 1.10Not started

    Counting principles and the extended binomial theorem

    Higher Level extends counting beyond the coefficients used in ordinary binomial expansion. Students distinguish arrangements from selections, use factorial notation efficiently and apply generalized binomial expansions when the exponent is not a non-negative integer.

    165 min estimated

    Topic 1.10 progress0%
    Learning objectives
    • Use systematic counting principles to determine the number of possible outcomes without double-counting.
    • Distinguish permutations, where order matters, from combinations, where order does not matter.
    • Use factorial notation and calculate nPr and nCr values analytically or with technology where appropriate.
    • Use the generalized binomial expansion for rational, fractional or negative exponents and identify a required coefficient or term.
    • Recognize the interval/condition under which an infinite binomial expansion is valid.
    Key formulas

    Fundamental counting principle

    N=m1m2⋯mkN=m_1m_2\cdots m_k

    Sum rule for disjoint cases

    N=N1+N2+⋯+NkN=N_1+N_2+\cdots+N_k

    Factorial definition

    n!=n(n−1)(n−2)⋯2⋅1n!=n(n-1)(n-2)\cdots2\cdot1

    Zero factorial

    0!=10!=1

    Useful cancellation

    n!(n−r)!=n(n−1)⋯(n−r+1)\frac{n!}{(n-r)!}=n(n-1)\cdots(n-r+1)

    Arrange all distinct objects

    N=n!N=n!

    Permutations of r from n

    nPr=n!(n−r)!{}_{n}P_r=\frac{n!}{(n-r)!}

    Combination formula

    (nr)=nCr=n!r!(n−r)!\binom{n}{r}={}_{n}C_r=\frac{n!}{r!(n-r)!}

    Combination symmetry

    (nr)=(nn−r)\binom{n}{r}=\binom{n}{n-r}

    Permutation-combination link

    nPr=r!(nr){}_{n}P_r=r!\binom{n}{r}

    Arrangements with identical objects

    N=n!n1!n2!⋯nk!N=\frac{n!}{n_1!n_2!\cdots n_k!}

    Sequences with repetition allowed

    N=krN=k^r

    Complement count

    N(valid)=N(all)−N(forbidden)N(\text{valid})=N(\text{all})-N(\text{forbidden})

    Finite binomial theorem

    (a+b)n=∑r=0n(nr)an−rbr(a+b)^n=\sum\limits_{r=0}^{n}\binom{n}{r}a^{n-r}b^r

    Finite general term

    Tr+1=(nr)an−rbrT_{r+1}=\binom{n}{r}a^{n-r}b^r

    Pascal identity

    (nr)=(n−1r−1)+(n−1r)\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}

    Generalized binomial coefficient

    (αr)=α(α−1)⋯(α−r+1)r!\binom{\alpha}{r}=\frac{\alpha(\alpha-1)\cdots(\alpha-r+1)}{r!}

    Extended binomial theorem

    (1+x)α=∑r=0∞(αr)xr,∣x∣<1(1+x)^\alpha=\sum\limits_{r=0}^{\infty}\binom{\alpha}{r}x^r,\qquad |x|<1

    First four terms

    (1+x)α=1+αx+α(α−1)2!x2+α(α−1)(α−2)3!x3+⋯(1+x)^\alpha=1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2+\frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3+\cdots

    Negative upper coefficient

    (−mr)=(−1)r(m+r−1r)\binom{-m}{r}=(-1)^r\binom{m+r-1}{r}

    Normalized binomial

    (a+bx)α=aα(1+bax)α(a+bx)^\alpha=a^\alpha\left(1+\frac{b}{a}x\right)^\alpha

    Resulting validity condition

    ∣bax∣<1\left|\frac{b}{a}x\right|<1

    Extended general term

    Tr+1=(αr)urT_{r+1}=\binom{\alpha}{r}u^r

    General term after normalization

    Tr+1=aα(αr)(bax)rT_{r+1}=a^\alpha\binom{\alpha}{r}\left(\frac{b}{a}x\right)^r

    Standard convergence condition

    ∣x∣<1|x|<1

    Substituted convergence condition

    ∣g(x)∣<1|g(x)|<1

    Truncated approximation

    (1+x)α≈∑r=0k(αr)xr(1+x)^\alpha\approx\sum\limits_{r=0}^{k}\binom{\alpha}{r}x^r

    Generalized coefficient recurrence

    (αr+1)=(αr)α−rr+1\binom{\alpha}{r+1}=\binom{\alpha}{r}\frac{\alpha-r}{r+1}

    Plausibility bound for a restriction

    0≤N(valid)≤N(all)0\le N(\text{valid})\le N(\text{all})

    Counting decision

    order matters⇒nPr,order does not matter⇒(nr)\text{order matters}\Rightarrow {}_{n}P_r,\qquad \text{order does not matter}\Rightarrow\binom{n}{r}

    Extended-expansion checklist

    (a+bx)α=aα(1+u)α,u=bax,∣u∣<1(a+bx)^\alpha=a^\alpha(1+u)^\alpha,\quad u=\frac{b}{a}x,\quad |u|<1

    Ordered selection

    nPr=n!(n−r)!{}_{n}P_r=\frac{n!}{(n-r)!}

    Unordered selection

    (nr)=n!r!(n−r)!\binom{n}{r}=\frac{n!}{r!(n-r)!}

    Factorial

    n!=n(n−1)⋯2⋅1,0!=1n!=n(n-1)\cdots2\cdot1,\qquad 0!=1

    Permutation and combination

    nPr=n!(n−r)!,(nr)=n!r!(n−r)!{}_{n}P_r=\frac{n!}{(n-r)!},\qquad \binom{n}{r}=\frac{n!}{r!(n-r)!}

    Repeated objects

    N=n!n1!n2!⋯nk!N=\frac{n!}{n_1!n_2!\cdots n_k!}

    Extended binomial theorem

    (1+u)α=∑r=0∞(αr)ur,∣u∣<1(1+u)^\alpha=\sum\limits_{r=0}^{\infty}\binom{\alpha}{r}u^r,\qquad |u|<1

    Generalized coefficient

    (αr)=α(α−1)⋯(α−r+1)r!\binom{\alpha}{r}=\frac{\alpha(\alpha-1)\cdots(\alpha-r+1)}{r!}
    Topic 1.11Not started

    Partial fractions

    Partial fractions reverse the process of combining rational expressions. The technique is important because a complicated rational function can often be rewritten as a sum of simpler fractions that is easier to integrate, simplify or analyze.

    155 min estimated

    Topic 1.11 progress0%
    Learning objectives
    • Check whether a rational expression is proper before decomposing it; use polynomial division first if necessary.
    • Factor the denominator and choose a decomposition that matches its factor structure.
    • Solve for unknown constants by equating coefficients or substituting convenient values.
    • Use the decomposition later in algebraic simplification and integration problems.
    Key formulas

    Basic decomposition identity

    P(x)Q(x)=∑k=1mNk(x)Fk(x)\frac{P(x)}{Q(x)}=\sum\limits_{k=1}^{m}\frac{N_k(x)}{F_k(x)}

    Polynomial division form

    P(x)=Q(x)S(x)+R(x)P(x)=Q(x)S(x)+R(x)

    Improper rational expression

    P(x)Q(x)=S(x)+R(x)Q(x)\frac{P(x)}{Q(x)}=S(x)+\frac{R(x)}{Q(x)}

    Remainder condition

    degree⁡R<degree⁡Q\operatorname{degree}R<\operatorname{degree}Q

    Repeated linear factor template

    A1x−a+A2(x−a)2+⋯+Am(x−a)m\frac{A_1}{x-a}+\frac{A_2}{(x-a)^2}+\cdots+\frac{A_m}{(x-a)^m}

    Irreducible quadratic template

    Bx+Cax2+bx+c\frac{Bx+C}{ax^2+bx+c}

    Quadratic irreducibility test

    b2−4ac<0b^2-4ac<0

    Two distinct linear factors

    P(x)(x−a)(x−b)=Ax−a+Bx−b\frac{P(x)}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}

    Cover-up coefficient

    A=P(x)Q(x)(x−a)∣x=aA=\left.\frac{P(x)}{\frac{Q(x)}{(x-a)}}\right|_{x=a}

    Repeated linear decomposition

    P(x)(x−a)mR(x)=A1x−a+⋯+Am(x−a)m+terms from R(x)\frac{P(x)}{(x-a)^mR(x)}=\frac{A_1}{x-a}+\cdots+\frac{A_m}{(x-a)^m}+\text{terms from }R(x)

    Highest-power coefficient

    Am=P(x)R(x)∣x=aA_m=\left.\frac{P(x)}{R(x)}\right|_{x=a}

    Irreducible quadratic contribution

    Bx+Cax2+bx+c\frac{Bx+C}{ax^2+bx+c}

    Repeated quadratic contribution

    ∑j=1mBjx+Cjq(x)j\sum\limits_{j=1}^{m}\frac{B_jx+C_j}{q(x)^j}

    Coefficient at a simple root

    A=lim⁡x→a(x−a)P(x)Q(x)A=\lim_{x\to a}(x-a)\frac{P(x)}{Q(x)}

    Highest repeated coefficient

    Am=lim⁡x→a(x−a)mP(x)Q(x)A_m=\lim_{x\to a}(x-a)^m\frac{P(x)}{Q(x)}

    Polynomial coefficient identity

    anxn+⋯+a0=bnxn+⋯+b0⇒ak=bka_nx^n+\cdots+a_0=b_nx^n+\cdots+b_0\Rightarrow a_k=b_k

    Reconstruction check

    ∑k=1mNk(x)Fk(x)=∑k=1mNk(x) Q(x)Fk(x)Q(x)\sum\limits_{k=1}^{m}\frac{N_k(x)}{F_k(x)}=\frac{\frac{\sum\limits_{k=1}^{m}N_k(x)\,Q(x)}{F_k(x)}}{Q(x)}

    Linear-factor integral

    ∫Ax−a dx=Aln⁡∣x−a∣+C\int\frac{A}{x-a}\,dx=A\ln|x-a|+C

    Basic telescoping decomposition

    1n(n+1)=1n−1n+1\frac1{n(n+1)}=\frac1n-\frac1{n+1}

    Complete workflow

    divide→factor→template→clear→solve→verify\text{divide}\to\text{factor}\to\text{template}\to\text{clear}\to\text{solve}\to\text{verify}

    Reasonableness identity

    P(x)=?∑k=1mNk(x)Q(x)Fk(x)P(x)\stackrel{?}{=}\sum\limits_{k=1}^{m}N_k(x)\frac{Q(x)}{F_k(x)}

    Distinct linear template

    Ax−a+Bx−b\frac{A}{x-a}+\frac{B}{x-b}

    Repeated linear template

    ∑j=1mAj(x−a)j\sum\limits_{j=1}^{m}\frac{A_j}{(x-a)^j}

    Quadratic template

    Bx+Cax2+bx+c\frac{Bx+C}{ax^2+bx+c}

    Improper-expression step

    PQ=S+RQ,degree⁡R<degree⁡Q\frac{P}{Q}=S+\frac{R}{Q},\qquad \operatorname{degree}R<\operatorname{degree}Q

    Distinct linear factors

    P(x)(x−a)(x−b)=Ax−a+Bx−b\frac{P(x)}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}

    Repeated linear factor

    ∑j=1mAj(x−a)j\sum\limits_{j=1}^{m}\frac{A_j}{(x-a)^j}

    Repeated irreducible quadratic

    ∑j=1mBjx+Cjq(x)j\sum\limits_{j=1}^{m}\frac{B_jx+C_j}{q(x)^j}
    Topic 1.12Not started

    Complex numbers in Cartesian form

    Complex numbers extend the real number system so that equations such as x2+1=0x^{2} + 1 = 0 have solutions. Students work algebraically with a + bi form and geometrically with points and vectors in the Argand plane.

    160 min estimated

    Topic 1.12 progress0%
    Learning objectives
    • Use i2=−1i^{2} = -1 to simplify powers of i and arithmetic involving complex numbers.
    • Identify the real part, imaginary part, conjugate, modulus and argument of a complex number.
    • Add, subtract, multiply and divide complex numbers in Cartesian form.
    • Represent complex numbers on an Argand diagram and interpret modulus as distance from the origin.
    • Use the conjugate to simplify quotients and connect complex roots with polynomial equations.
    Key formulas

    Imaginary unit

    i2=−1i^2=-1

    Cartesian form

    z=a+biz=a+bi

    Set of complex numbers

    C={a+bi:a,b∈R}\mathbb{C}=\{a+bi:a,b\in\mathbb{R}\}

    Real part

    Re⁡(a+bi)=a\operatorname{Re}(a+bi)=a

    Imaginary part

    Im⁡(a+bi)=b\operatorname{Im}(a+bi)=b

    Real-number condition

    z∈R  ⟺  Im⁡(z)=0z\in\mathbb{R}\iff\operatorname{Im}(z)=0

    Complex equality

    a+bi=c+di  ⟺  a=c and b=da+bi=c+di\iff a=c\text{ and }b=d

    Addition

    (a+bi)+(c+di)=(a+c)+(b+d)i(a+bi)+(c+di)=(a+c)+(b+d)i

    Subtraction

    (a+bi)−(c+di)=(a−c)+(b−d)i(a+bi)-(c+di)=(a-c)+(b-d)i

    Four-power cycle

    i0=1,i1=i,i2=−1,i3=−i,i4=1i^0=1,\quad i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1

    Periodicity

    in+4=ini^{n+4}=i^n

    Inverse of i

    i−1=−ii^{-1}=-i

    Complex multiplication

    (a+bi)(c+di)=(ac−bd)+(ad+bc)i(a+bi)(c+di)=(ac-bd)+(ad+bc)i

    Complex square

    (a+bi)2=(a2−b2)+2abi(a+bi)^2=(a^2-b^2)+2abi

    Conjugate

    a+bi‾=a−bi\overline{a+bi}=a-bi

    Sum with conjugate

    z+z‾=2Re⁡(z)z+\overline z=2\operatorname{Re}(z)

    Difference from conjugate

    z−z‾=2iIm⁡(z)z-\overline z=2i\operatorname{Im}(z)

    Conjugate product

    zz‾=a2+b2z\overline z=a^2+b^2

    Conjugate of a product

    zw‾=z‾ w‾\overline{zw}=\overline z\,\overline w

    Complex quotient

    a+bic+di=(ac+bd)+(bc−ad)ic2+d2\frac{a+bi}{c+di}=\frac{(ac+bd)+(bc-ad)i}{c^2+d^2}

    Complex reciprocal

    1a+bi=a−bia2+b2\frac1{a+bi}=\frac{a-bi}{a^2+b^2}

    Modulus

    ∣a+bi∣=a2+b2|a+bi|=\sqrt{a^2+b^2}

    Modulus squared

    ∣z∣2=zz‾|z|^2=z\overline z

    Distance

    d(z1,z2)=∣z1−z2∣d(z_1,z_2)=|z_1-z_2|

    Product modulus

    ∣zw∣=∣z∣∣w∣|zw|=|z||w|

    Triangle inequality

    ∣z+w∣≤∣z∣+∣w∣|z+w|\le |z|+|w|

    Argand coordinate

    z=a+bi⟷(a,b)z=a+bi\longleftrightarrow(a,b)

    Conjugate reflection

    a+bi⟷a−bia+bi\longleftrightarrow a-bi

    Displacement

    z1−z2z_1-z_2

    Square in Cartesian form

    (a+bi)2=(a2−b2)+2abi(a+bi)^2=(a^2-b^2)+2abi

    Quadratic formula

    z=−b±b2−4ac2az=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

    Conjugate factor pair

    (x−(a+bi))(x−(a−bi))=(x−a)2+b2(x-(a+bi))(x-(a-bi))=(x-a)^2+b^2

    Quadratic from conjugate roots

    x2−2ax+(a2+b2)=0x^2-2ax+(a^2+b^2)=0

    Circle locus

    ∣z−c∣=r|z-c|=r

    Equidistance locus

    ∣z−z1∣=∣z−z2∣|z-z_1|=|z-z_2|

    Universal arithmetic check

    computed answer→substitutionoriginal equation\text{computed answer}\xrightarrow{\text{substitution}}\text{original equation}

    Core Cartesian workflow

    expand→use i2=−1→collect real and imaginary parts\text{expand}\to\text{use }i^2=-1\to\text{collect real and imaginary parts}

    Conjugate

    z‾=a−bi\overline z=a-bi

    Modulus

    ∣z∣=a2+b2|z|=\sqrt{a^2+b^2}

    Division

    a+bic+di=(ac+bd)+(bc−ad)ic2+d2\frac{a+bi}{c+di}=\frac{(ac+bd)+(bc-ad)i}{c^2+d^2}
    Topic 1.13Not started

    Polar, cis and exponential forms of complex numbers

    Polar form separates a complex number into its modulus and direction. It is especially efficient for multiplication, division, powers and roots, where geometric rotation and scaling become visible.

    160 min estimated

    Topic 1.13 progress0%
    Learning objectives
    • Convert between Cartesian and polar/exponential forms.
    • Choose an argument consistent with the point’s quadrant and the required argument convention.
    • Multiply and divide complex numbers by combining moduli and arguments.
    • Interpret multiplication geometrically as a scaling and rotation in the complex plane.
    Key formulas

    Modulus

    r=∣z∣=x2+y2r=|z|=\sqrt{x^2+y^2}

    Cartesian projections

    x=rcos⁡θ,y=rsin⁡θx=r\cos\theta,\qquad y=r\sin\theta

    Tangent relation

    tan⁡θ=yx\tan\theta=\frac{y}{x}

    Polar form

    z=r(cos⁡θ+isin⁡θ)z=r(\cos\theta+i\sin\theta)

    Equivalent arguments

    θk=θ+2kπ,k∈Z\theta_k=\theta+2k\pi,\qquad k\in\mathbb Z

    Cartesian reconstruction

    z=rcos⁡θ+irsin⁡θz=r\cos\theta+i r\sin\theta

    Cis definition

    cis⁡θ=cos⁡θ+isin⁡θ\operatorname{cis}\theta=\cos\theta+i\sin\theta

    Cis form

    z=rcis⁡θz=r\operatorname{cis}\theta

    Cis periodicity

    cis⁡(θ+2kπ)=cis⁡θ\operatorname{cis}(\theta+2k\pi)=\operatorname{cis}\theta

    Euler's relation

    eiθ=cos⁡θ+isin⁡θe^{i\theta}=\cos\theta+i\sin\theta

    Exponential form

    z=reiθz=re^{i\theta}

    Equivalent notation

    reiθ=rcis⁡θre^{i\theta}=r\operatorname{cis}\theta

    Principal-argument convention

    −π<Arg⁡z≤π-\pi<\operatorname{Arg}z\le\pi

    General arguments

    arg⁡z=Arg⁡z+2kπ,k∈Z\arg z=\operatorname{Arg}z+2k\pi,\qquad k\in\mathbb Z

    Cartesian-to-polar modulus

    r=x2+y2r=\sqrt{x^2+y^2}

    Cartesian-to-polar argument

    θ=atan2⁡(y,x)\theta=\operatorname{atan2}(y,x)

    Equivalent output forms

    r(cos⁡θ+isin⁡θ)=rcis⁡θ=reiθr(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta=re^{i\theta}

    Real component

    x=rcos⁡θx=r\cos\theta

    Imaginary component

    y=rsin⁡θy=r\sin\theta

    Polar-to-Cartesian conversion

    rcis⁡θ=rcos⁡θ+irsin⁡θr\operatorname{cis}\theta=r\cos\theta+i r\sin\theta

    Product in cis form

    z1z2=r1r2cis⁡(θ1+θ2)z_1z_2=r_1r_2\operatorname{cis}(\theta_1+\theta_2)

    Product in exponential form

    z1z2=r1r2ei(θ1+θ2)z_1z_2=r_1r_2e^{i(\theta_1+\theta_2)}

    Product modulus

    ∣z1z2∣=∣z1∣∣z2∣|z_1z_2|=|z_1||z_2|

    Quotient in cis form

    z1z2=r1r2cis⁡(θ1−θ2)\frac{z_1}{z_2}=\frac{r_1}{r_2}\operatorname{cis}(\theta_1-\theta_2)

    Reciprocal

    1rcis⁡θ=1rcis⁡(−θ)\frac1{r\operatorname{cis}\theta}=\frac1r\operatorname{cis}(-\theta)

    Conjugate in polar form

    rcis⁡θ‾=rcis⁡(−θ)\overline{r\operatorname{cis}\theta}=r\operatorname{cis}(-\theta)

    Quotient modulus

    ∣z1z2∣=∣z1∣∣z2∣\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}

    Scaling and rotation

    z↦kcis⁡ϕ zz\mapsto k\operatorname{cis}\phi\,z

    Modulus transformation

    ∣kcis⁡ϕ z∣=k∣z∣|k\operatorname{cis}\phi\,z|=k|z|

    Argument transformation

    arg⁡(kcis⁡ϕ z)=arg⁡z+ϕ(mod2π)\arg(k\operatorname{cis}\phi\,z)=\arg z+\phi\pmod{2\pi}

    Axis arguments

    Arg⁡(a)=0 (a>0),Arg⁡(a)=π (a<0)\operatorname{Arg}(a)=0\ (a>0),\quad \operatorname{Arg}(a)=\pi\ (a<0)

    Imaginary-axis arguments

    Arg⁡(bi)=π2 (b>0),Arg⁡(bi)=−π2 (b<0)\operatorname{Arg}(bi)=\frac{\pi}{2}\ (b>0),\quad \operatorname{Arg}(bi)=-\frac{\pi}{2}\ (b<0)

    Positive-modulus conversion

    −rcis⁡θ=rcis⁡(θ+π)-r\operatorname{cis}\theta=r\operatorname{cis}(\theta+\pi)

    Operation choice

    add/subtract→Cartesian,multiply/divide→polar\text{add/subtract}\to\text{Cartesian},\qquad \text{multiply/divide}\to\text{polar}

    Quadrant-aware argument

    θ=atan2⁡(y,x)\theta=\operatorname{atan2}(y,x)

    Post-operation check

    ∣z1z2∣=∣z1∣∣z2∣,arg⁡(z1z2)=arg⁡z1+arg⁡z2(mod2π)|z_1z_2|=|z_1||z_2|,\qquad \arg(z_1z_2)=\arg z_1+\arg z_2\pmod{2\pi}

    Principal normalization

    θ↦θ+2kπ∈(−π,π]\theta\mapsto\theta+2k\pi\in(-\pi,\pi]

    Conversion-operation workflow

    identify form→find r,θ→operate→normalize→verify\text{identify form}\to\text{find }r,\theta\to\text{operate}\to\text{normalize}\to\text{verify}

    Modulus

    r=x2+y2r=\sqrt{x^2+y^2}

    Product

    z1z2=r1r2cis⁡(θ1+θ2)z_1z_2=r_1r_2\operatorname{cis}(\theta_1+\theta_2)

    Quotient

    z1z2=r1r2cis⁡(θ1−θ2)\frac{z_1}{z_2}=\frac{r_1}{r_2}\operatorname{cis}(\theta_1-\theta_2)
    Topic 1.14Not started

    Conjugate roots and De Moivre’s theorem

    De Moivre’s theorem makes powers and roots of complex numbers systematic. Students also use the conjugate-root property of real-coefficient polynomials to connect complex solutions back to algebraic equations.

    170 min estimated

    Topic 1.14 progress0%
    Learning objectives
    • Use the fact that non-real roots of a polynomial with real coefficients occur in conjugate pairs.
    • Apply De Moivre’s theorem to calculate powers exactly.
    • Find all nth roots of a complex number and display their regular angular spacing on an Argand diagram.
    • Use polar form to solve equations involving powers of complex numbers.
    Key formulas

    Complex conjugate

    a+bi‾=a−bi\overline{a+bi}=a-bi

    Conjugate product

    zz‾=∣z∣2=a2+b2z\overline z=|z|^2=a^2+b^2

    Conjugate in polar form

    rcis⁡θ‾=rcis⁡(−θ)\overline{r\operatorname{cis}\theta}=r\operatorname{cis}(-\theta)

    Conjugation of a real polynomial

    p(z‾)=p(z)‾p(\overline z)=\overline{p(z)}

    Conjugate-root implication

    p(z)=0  ⟹  p(z‾)=0p(z)=0\implies p(\overline z)=0

    Conjugate factor pair

    (x−(a+bi))(x−(a−bi))=(x−a)2+b2(x-(a+bi))(x-(a-bi))=(x-a)^2+b^2

    Expanded real quadratic

    x2−2ax+(a2+b2)x^2-2ax+(a^2+b^2)

    Repeated conjugate factor

    ((x−a)2+b2)m\left((x-a)^2+b^2\right)^m

    De Moivre's theorem

    (rcis⁡θ)n=rncis⁡(nθ)(r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta)

    Unit-modulus form

    (cos⁡θ+isin⁡θ)n=cos⁡(nθ)+isin⁡(nθ)(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta)

    Exponential equivalent

    (reiθ)n=rneinθ(re^{i\theta})^n=r^ne^{in\theta}

    Power modulus

    ∣zn∣=∣z∣n|z^n|=|z|^n

    Power argument

    arg⁡(zn)=narg⁡z(mod2π)\arg(z^n)=n\arg z\pmod{2\pi}

    Triple-angle cosine

    cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos3\theta=4\cos^3\theta-3\cos\theta

    Triple-angle sine

    sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin3\theta=3\sin\theta-4\sin^3\theta

    Nth-root formula

    zk=R1ncis⁡(θ+2kπn),k=0,1,…,n−1z_k=R^{\frac{1}{n}}\operatorname{cis}\left(\frac{\theta+2k\pi}{n}\right),\qquad k=0,1,\ldots,n-1

    Root modulus

    ∣zk∣=R1n|z_k|=R^{\frac{1}{n}}

    Root angular spacing

    Δθ=2πn\Delta\theta=\frac{2\pi}{n}

    Root angles

    ϕk=θ+2kπn\phi_k=\frac{\theta+2k\pi}{n}

    Root check

    (zk)n=w(z_k)^n=w

    Root circle

    ∣z∣=R1n|z|=R^{\frac{1}{n}}

    Consecutive-root rotation

    zk+1=zkcis⁡2πnz_{k+1}=z_k\operatorname{cis}\frac{2\pi}{n}

    Root sum

    ∑k=0n−1zk=0,n>1\sum\limits_{k=0}^{n-1}z_k=0,\qquad n>1

    Nth roots of unity

    ωk=cis⁡2kπn,k=0,1,…,n−1\omega_k=\operatorname{cis}\frac{2k\pi}{n},\qquad k=0,1,\ldots,n-1

    Root-of-unity equation

    zn−1=0z^n-1=0

    Primitive-root condition

    gcd⁡(k,n)=1\gcd(k,n)=1

    Sum of nth roots of unity

    ∑k=0n−1ωk=0,n>1\sum\limits_{k=0}^{n-1}\omega_k=0,\qquad n>1

    Factor theorem

    p(α)=0  ⟺  (x−α) is a factor of p(x)p(\alpha)=0\iff(x-\alpha)\text{ is a factor of }p(x)

    Root-count rule

    deg⁡p=n  ⟹  n complex roots counting multiplicity\deg p=n\implies n\text{ complex roots counting multiplicity}

    Shifted power substitution

    (z−a)n=wwith u=z−a(z-a)^n=w\quad\text{with }u=z-a

    Root-family completeness

    k=0,1,…,n−1k=0,1,\ldots,n-1

    Numerical root check

    ∣p(zk)∣≈0|p(z_k)|\approx0

    Geometric root check

    ∣zk∣=R1n,arg⁡zk+1−arg⁡zk=2πn|z_k|=R^{\frac{1}{n}},\qquad \arg z_{k+1}-\arg z_k=\frac{2\pi}{n}

    Power checklist

    rcis⁡θ→nrncis⁡(nθ)r\operatorname{cis}\theta\xrightarrow{n}r^n\operatorname{cis}(n\theta)

    Root checklist

    Rcis⁡θ→nth rootsR1ncis⁡(θ+2kπn)R\operatorname{cis}\theta\xrightarrow{n\text{th roots}}R^{\frac{1}{n}}\operatorname{cis}\left(\frac{\theta+2k\pi}{n}\right)

    Complete workflow

    identify theorem→polar form→power or roots→complete set→verify\text{identify theorem}\to\text{polar form}\to\text{power or roots}\to\text{complete set}\to\text{verify}

    Conjugate-root factor

    (x−(a+bi))(x−(a−bi))=(x−a)2+b2(x-(a+bi))(x-(a-bi))=(x-a)^2+b^2

    Nth roots

    zk=R1ncis⁡(θ+2kπn)z_k=R^{\frac{1}{n}}\operatorname{cis}\left(\frac{\theta+2k\pi}{n}\right)

    Roots of unity

    ωk=cis⁡2kπn\omega_k=\operatorname{cis}\frac{2k\pi}{n}
    Topic 1.15Not started

    Proof by induction, contradiction and counterexample

    Higher Level expects students to communicate proof with explicit logical structure. Induction establishes a statement for every integer in a sequence of cases; contradiction assumes the opposite and derives an impossibility; a counterexample disproves a universal statement by finding one valid exception.

    175 min estimated

    Topic 1.15 progress0%
    Learning objectives
    • Write a complete induction proof with a base case, inductive hypothesis and inductive step.
    • Use contradiction when assuming the negation of a statement leads to a logical impossibility.
    • For the current syllabus, use a counterexample to show that a universal claim is false.
    • Distinguish proof from numerical evidence: checking many cases is not the same as proving all cases.
    Key formulas

    Implication

    P  ⟹  QP\implies Q

    Universal statement

    ∀x∈D, P(x)\forall x\in D,\ P(x)

    Existential statement

    ∃x∈D such that P(x)\exists x\in D\text{ such that }P(x)

    Contradiction form for an implication

    P∧¬Q  ⟹  ⊥P\land\neg Q\implies\bot

    Disproving a universal claim

    ¬(∀x∈D,P(x))  ⟺  ∃x∈D with ¬P(x)\neg(\forall x\in D, P(x))\iff\exists x\in D\text{ with }\neg P(x)

    Induction principle

    P(n0)∧(∀k≥n0, P(k)  ⟹  P(k+1))  ⟹  ∀n≥n0, P(n)P(n_0)\land\big(\forall k\ge n_0,\ P(k)\implies P(k+1)\big)\implies\forall n\ge n_0,\ P(n)

    Base-case instruction

    n=n0:verify P(n0)n=n_0:\quad\text{verify }P(n_0)

    Inductive hypothesis

    P(k) is assumed for an arbitrary k≥n0P(k)\text{ is assumed for an arbitrary }k\ge n_0

    Divisibility hypothesis

    d∣Ak  ⟺  Ak=dm for some m∈Zd\mid A_k\iff A_k=dm\text{ for some }m\in\mathbb Z

    Sum extension

    ∑r=1k+1ar=∑r=1kar+ak+1\sum\limits_{r=1}^{k+1}a_r=\sum\limits_{r=1}^{k}a_r+a_{k+1}

    Inductive implication

    P(k)  ⟹  P(k+1)P(k)\implies P(k+1)

    Natural-number sum

    ∑r=1nr=n(n+1)2\sum\limits_{r=1}^{n}r=\frac{n(n+1)}2

    Odd-number sum

    ∑r=1n(2r−1)=n2\sum\limits_{r=1}^{n}(2r-1)=n^2

    Finite geometric sum

    ∑r=0nqr=qn+1−1q−1\sum\limits_{r=0}^{n}q^r=\frac{q^{n+1}-1}{q-1}

    Divisibility definition

    d∣A  ⟺  A=dm for some m∈Zd\mid A\iff A=dm\text{ for some }m\in\mathbb Z

    Useful power rearrangement

    ak+1−1=a(ak−1)+(a−1)a^{k+1}-1=a(a^k-1)+(a-1)

    One-step recurrence proof

    uk+1=f(uk)→P(k)f(g(k))=g(k+1)u_{k+1}=f(u_k)\xrightarrow{P(k)}f(g(k))=g(k+1)

    Strong induction hypothesis

    P(n0),P(n0+1),…,P(k)P(n_0),P(n_0+1),\ldots,P(k)

    Example inequality chain

    2k+1=2⋅2k≥2(k+1)≥k+22^{k+1}=2\cdot2^k\ge2(k+1)\ge k+2

    Contradiction schema

    ¬S  ⟹  ⊥∴S\neg S\implies\bot\quad\therefore\quad S

    Implication by contradiction

    P∧¬Q  ⟹  ⊥∴P  ⟹  QP\land\neg Q\implies\bot\quad\therefore\quad P\implies Q

    Even integer

    n=2m,m∈Zn=2m,\qquad m\in\mathbb Z

    Odd integer

    n=2m+1,m∈Zn=2m+1,\qquad m\in\mathbb Z

    Euclid construction

    N=p1p2⋯pm+1N=p_1p_2\cdots p_m+1

    Counterexample to implication

    P(c) true and Q(c) falseP(c)\text{ true and }Q(c)\text{ false}

    Universal negation

    ¬(∀x∈D,P(x))  ⟺  ∃c∈D,¬P(c)\neg(\forall x\in D,P(x))\iff\exists c\in D,\neg P(c)

    Induction checklist

    P(n0);P(k) assumed;P(k+1) proved;conclude for all n≥n0P(n_0);\quad P(k)\text{ assumed};\quad P(k+1)\text{ proved};\quad\text{conclude for all }n\ge n_0

    Contradiction checklist

    assume negation→deduce impossibility→reject negation\text{assume negation}\to\text{deduce impossibility}\to\text{reject negation}

    Proof workflow

    identify quantifier→choose method→justify every step→state conclusion\text{identify quantifier}\to\text{choose method}\to\text{justify every step}\to\text{state conclusion}

    Induction structure

    P(n0);P(k)  ⟹  P(k+1);∴ P(n) for all n≥n0P(n_0);\quad P(k)\implies P(k+1);\quad\therefore\ P(n)\text{ for all }n\ge n_0

    Contradiction structure

    ¬S  ⟹  ⊥∴S\neg S\implies\bot\quad\therefore\quad S

    Counterexample structure

    c∈D,P(c) is false∴∀x∈D,P(x) is falsec\in D,\quad P(c)\text{ is false}\quad\therefore\quad\forall x\in D,P(x)\text{ is false}
    Topic 1.16Not started

    Systems of linear equations

    Students solve and interpret systems of two or three linear equations using algebra, matrices, Gaussian elimination and technology. They distinguish unique, inconsistent and dependent systems, including parameter cases, and connect row structure with the geometry of lines and planes.

    175 min estimated

    Topic 1.16 progress0%
    Learning objectives
    • Explain a system of linear equations and verify an ordered pair or triple in every original equation.
    • Solve systems with two unknowns by substitution and elimination using exact arithmetic.
    • Solve systems with three unknowns by systematic elimination and back-substitution.
    • Represent a system as an augmented matrix and perform valid elementary row operations.
    • Use Gaussian elimination or reduced row-echelon form to solve and classify a system.
    • Distinguish unique solutions, no solutions and infinitely many solutions from equations, row-echelon form and geometry.
    • Identify consistent, inconsistent, independent and dependent systems and communicate free parameters correctly.
    • Analyze parameter-based systems without dividing by an expression that may be zero.
    • Use a GDC to solve or check systems while preserving exact setup, interpretation and validation.
    • Model real situations with linear systems, interpret the solution with units and test whether it is reasonable.
    Key formulas

    General two-variable linear system

    a1x+b1y=c1,a2x+b2y=c2a_1x+b_1y=c_1,\qquad a_2x+b_2y=c_2

    General three-variable linear equation

    ax+by+cz=dax+by+cz=d

    Substitution pattern

    x=g(y)⟹a g(y)+by=cx=g(y)\quad\Longrightarrow\quad a\,g(y)+by=c

    Elimination combination

    λ(a1x+b1y=c1)+μ(a2x+b2y=c2)\lambda(a_1x+b_1y=c_1)+\mu(a_2x+b_2y=c_2)

    Three-equation model

    a1x+b1y+c1z=d1a2x+b2y+c2z=d2a3x+b3y+c3z=d3\begin{aligned}a_1x+b_1y+c_1z&=d_1\\a_2x+b_2y+c_2z&=d_2\\a_3x+b_3y+c_3z&=d_3\end{aligned}

    Matrix form

    Ax=bA\mathbf{x}=\mathbf{b}

    Augmented matrix

    [A∣b][A\mid\mathbf{b}]

    Two-variable coefficient matrix

    A=[a1b1a2b2]A=\begin{bmatrix}a_1&b_1\\a_2&b_2\end{bmatrix}

    Row interchange

    Ri↔RjR_i\leftrightarrow R_j

    Nonzero row scaling

    Ri→λRi,λ≠0R_i\to\lambda R_i,\qquad\lambda\ne0

    Row replacement

    Ri→Ri+λRjR_i\to R_i+\lambda R_j

    Row-echelon structure

    [∗∗∗0∗∗00∗]\begin{bmatrix}*&*&*\\0&*&*\\0&0&*\end{bmatrix}

    Back-substitution direction

    z⟶y⟶xz\longrightarrow y\longrightarrow x

    Consistency criterion

    rank⁡(A)=rank⁡([A∣b])\operatorname{rank}(A)=\operatorname{rank}([A\mid\mathbf b])

    Unique-solution rank condition

    rank⁡(A)=rank⁡([A∣b])=n\operatorname{rank}(A)=\operatorname{rank}([A\mid\mathbf b])=n

    Infinite-solution rank condition

    rank⁡(A)=rank⁡([A∣b])<n\operatorname{rank}(A)=\operatorname{rank}([A\mid\mathbf b])<n

    No-solution rank condition

    rank⁡(A)<rank⁡([A∣b])\operatorname{rank}(A)<\operatorname{rank}([A\mid\mathbf b])

    Two-line determinant

    D=a1b2−a2b1D=a_1b_2-a_2b_1

    Nonparallel-line condition

    D≠0⟹one intersectionD\ne0\quad\Longrightarrow\quad\text{one intersection}

    Two-by-two determinant

    det⁡A=∣a1b1a2b2∣=a1b2−a2b1\det A=\begin{vmatrix}a_1&b_1\\a_2&b_2\end{vmatrix}=a_1b_2-a_2b_1

    Unique 2 by 2 system

    det⁡A≠0\det A\ne0

    Parameter case trigger

    p(k)=0orp(k)≠0p(k)=0\quad\text{or}\quad p(k)\ne0

    Determinant case split

    det⁡A(k)≠0⇒unique;det⁡A(k)=0⇒inspect [A∣b]\det A(k)\ne0\Rightarrow\text{unique};\qquad \det A(k)=0\Rightarrow\text{inspect }[A\mid\mathbf b]

    Two-category total

    x+y=Tx+y=T

    Weighted-total model

    p1x+p2y=Pp_1x+p_2y=P

    Inverse-matrix solution

    x=A−1b,det⁡A≠0\mathbf{x}=A^{-1}\mathbf{b},\qquad\det A\ne0

    Residual vector

    r=Ax^−b\mathbf r=A\widehat{\mathbf x}-\mathbf b

    Solution verification

    Ax^=bA\widehat{\mathbf x}=\mathbf b

    System form

    Ax=bA\mathbf x=\mathbf b

    Allowed row replacement

    Ri→Ri+λRjR_i\to R_i+\lambda R_j

    Consistency test

    rank⁡(A)=rank⁡([A∣b])\operatorname{rank}(A)=\operatorname{rank}([A\mid\mathbf b])

    Unique solution test

    rank⁡(A)=rank⁡([A∣b])=n\operatorname{rank}(A)=\operatorname{rank}([A\mid\mathbf b])=n

    Infinite solutions test

    rank⁡(A)=rank⁡([A∣b])<n\operatorname{rank}(A)=\operatorname{rank}([A\mid\mathbf b])<n

    No solution test

    rank⁡(A)<rank⁡([A∣b])\operatorname{rank}(A)<\operatorname{rank}([A\mid\mathbf b])

    Two-by-two determinant test

    det⁡A=a1b2−a2b1\det A=a_1b_2-a_2b_1

    Inverse method condition

    x=A−1b,det⁡A≠0\mathbf x=A^{-1}\mathbf b,\qquad\det A\ne0
  2. Unit 2
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    Functions

    Functions are a central organizing idea in AA HL. The SL core develops function notation, transformations, inverses, quadratics and exponential/logarithmic behaviour; Higher Level extends this to general polynomials, more complicated rational functions, parity, inequalities and transformations that substantially change graph structure.

    16 topics

    Unit 2 progress0%
  3. Unit 3
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    Geometry and Trigonometry

    Geometry and Trigonometry is the largest HL topic by recommended teaching time. The SL core develops triangle trigonometry, radians, exact values, identities and circular functions; the HL extension adds reciprocal and inverse trigonometric functions, compound-angle identities and an extensive three-dimensional vector treatment of lines, planes, intersections and angles.

    18 topics

    Unit 3 progress0%
  4. Unit 4
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    Statistics and Probability

    AA HL statistics and probability begins with the same data, correlation, probability and distribution content studied at SL, then extends conditional probability with Bayes’ theorem and develops the theory of discrete and continuous random variables in greater depth.

    14 topics

    Unit 4 progress0%
  5. Unit 5
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    Calculus

    Calculus is the largest analytical component of AA HL. The SL core establishes differentiation, integration, optimization, motion and area; Higher Level develops rigorous limits, first-principles differentiation, l’Hôpital’s rule, implicit differentiation, inverse-trigonometric and reciprocal-trigonometric calculus, advanced integration, volumes, differential equations and Maclaurin series.

    19 topics

    Unit 5 progress0%